If a denotes the number of permutations of x | vedclass

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(B) Given that $a = {}^{x+2}P_{x+2} = (x+2)!$,$b = {}^{x}P_{11} = \frac{x!}{(x-11)!}$,and $c = {}^{x-11}P_{x-11} = (x-11)!$.Given the equation $a = 18

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$12$

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(B) Given that $a = {}^{x+2}P_{x+2} = (x+2)!$,$b = {}^{x}P_{11} = \frac{x!}{(x-11)!}$,and $c = {}^{x-11}P_{x-11} = (x-11)!$.Given the equation $a = 182bc$,we substitute the expressions:$(x+2)! = 182 	imes \frac{x!}{(x-11)!} 	imes (x-11)!$Simplifying the equation:$(x+2)! = 182 	imes x!$$(x+2)(x+1)x! = 182 	imes x!$Dividing both sides by $x!$ (where $x \ge 11$):$(x+2)(x+1) = 182$$x^2 + 3x + 2 = 182$$x^2 + 3x - 180 = 0$Factoring the quadratic equation:$(x+15)(x-12) = 0$Since $x$ must be a positive integer and $x \ge 11$,we have $x = 12$.

If $a$ denotes the number of permutations of $x + 2$ things taken all at a time,$b$ the number of permutations of $x$ things taken $11$ at a time,and $c$ the number of permutations of $x - 11$ things taken all at a time such that $a = 182bc$,then the value of $x$ is

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